Question

Given circle with centre $O$ and $BC\parallel ED$, $m\left(arcBC\right)={94}^0$, $m\left(arcED\right)={86}^0$, $\angle ADE=8^0$.

The m$\angle ECB$ is :

• A. ${43}^o$
• B. ${45}^o$
• C. ${75}^o$
• D. ${40}^o$

Answer 1

The correct option is B ${45}^o$

We have $\angle BOC={94}^{\circ },\angle EOD={86}^{\circ }$ and $\angle BOE={90}^{\circ }$. Hence, from angles around a point, we have $\angle COD={90}^{\circ }$

$\therefore$ in $△OEC,\angle COE=\angle COD+\angle DOE={176}^{\circ }$ and as $OC=OE$, $\angle OCE=\angle OEC=\frac{{180}^{\circ }-{176}^{\circ }}{2}=2^{\circ }$

Also, $\angle OCB=\angle OBC=\frac{{180}^{\circ }-{94}^{\circ }}{2}={43}^{\circ }$

$\therefore \angle ECB=\angle OCE+\angle OCB={45}^{\circ }$

This is the required answer.